# Complex sets: factoring into circle

How must $|z|=3|z-1|$ be factored so I end up with a circle, plugging in $z=x+iy$ seems to just up with square roots everywhere. Detailed steps is much appreciated, thanks!

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Square it up!.. – Berci Feb 6 '13 at 22:37
Have u tried its a mess – DJ_ Feb 6 '13 at 22:38
And use that $|z|^2=z \bar{z}$. – coffeemath Feb 6 '13 at 22:39
Yes, we can also calculate with complex numbers, using $|z|^2=z\bar z$: $$z\bar z=9(z-1)\overline{(z-1)}$$ which you prefer. – Berci Feb 6 '13 at 22:41

\begin{align*} z \overline{z} = 9 (z - 1)(\overline{z}-1) = 9 z \overline{z} - 9 z - 9 \overline{z} + 9 \\ 8 z \overline{z} - 9 z - 9 \overline{z} + 9 = 0 \end{align*} Now a circle is supposed to look like $$(z - a)(\overline{z} - \overline{a}) = r^2= z \overline{z} - a \overline{z} - \overline{a} z + |a|^2$$ So taking our old relation and diving by 8, we get \begin{align*} z \overline{z} - \frac{9}{8} z - \frac{9}{8} \overline{z} + \frac{9}{8} = 0 \\ z \overline{z} - \frac{9}{8} z - \frac{9}{8} \overline{z} + \frac{81}{64} = \frac{81}{64} - \frac{9}{8} = \frac{9}{64} \end{align*} So with some algebra we have a circle centered at $\frac{9}{8}$ with radius $\frac{3}{8}$.
For fun: you can view the locus of solutions $S$ as the inverse image of the unit circle of the fractional linear transformation $z \mapsto \frac{z}{3z-3}$. Conjugate points are sent to conjugate points and the center of $S$'s conjugate is $\infty$. Infinity maps to $1/3$ and the conjugate about the unit circle is $3$, so the inverse of that is $8/9$. Similarly I can find what is sent to 1 and that'll give me a radius.
Using $z=x+iy$, we get $$x^2+y^2=|z|^2=9|z-1|^2=9((x-1)^2+y^2).$$